If the coefficient of kinetic friction is .25, how much horizontal force is needed to pull each of the following masses along a rough desk at a constant speed?

a. 25 kg
b. 15 kg
c. 250 g

explain please!!

force friction=mu*m*g

at constant velocity, the pushing force is exactly equal to the force of friction, so
pushingforce=mu*mass*accelerationgravity

thanks!!!

To determine the amount of horizontal force needed to pull each of the masses along a rough desk at a constant speed, we need to use the equation for kinetic friction.

The equation for kinetic friction is:

Fk = μk * N

Where:
Fk is the force of kinetic friction,
μk is the coefficient of kinetic friction,
and N is the normal force.

The normal force is the force exerted by a surface perpendicular to the contact area. In this case, it is equal to the weight of the object.

a. For the 25 kg mass:
The normal force (N) is equal to the weight (W). The weight is the mass (m) multiplied by the acceleration due to gravity (g).

N = W = m * g = 25 kg * 9.8 m/s^2 = 245 N

Using the equation for kinetic friction:

Fk = μk * N = 0.25 * 245 N = 61.25 N

Therefore, you would need a horizontal force of 61.25 Newtons to pull the 25 kg mass along the rough desk at a constant speed.

b. For the 15 kg mass:
Similarly, calculate the normal force:

N = W = m * g = 15 kg * 9.8 m/s^2 = 147 N

Using the equation for kinetic friction:

Fk = μk * N = 0.25 * 147 N = 36.75 N

Therefore, you would need a horizontal force of 36.75 Newtons to pull the 15 kg mass along the rough desk at a constant speed.

c. For the 250 g mass:
Convert the mass to kilograms:

m = 250 g = 0.25 kg

Calculate the normal force:

N = W = m * g = 0.25 kg * 9.8 m/s^2 = 2.45 N

Using the equation for kinetic friction:

Fk = μk * N = 0.25 * 2.45 N = 0.6125 N

Therefore, you would need a horizontal force of 0.6125 Newtons to pull the 250 g mass along the rough desk at a constant speed.